Two point charges qA = 3 µC and qB= –3 µC are located 20 cm apart in a vacuum. (i) What is the electric field at the midpoint O of the line AB joining the two charges? (ii) If a negative test charge of magnitude 1.5 × 10–9 C is placed at this point, what is the force experienced by the test charge?
Distance between the two charges, AB=20cm ∴AO=OB=10cm Net electric field at point O=E Electric field at point O caused by +3 C charge, E1=4π∈0(AO)23×10−6=4π∈0(10×10−2)23×10−6N/C along OB Where, ∈0= Permittivity of free space 4π∈01=9×109Nm2C−2 Magnitude of electric field at point O caused by - 3 C charge, E2=∣∣∣∣∣4π∈0(OB)2−3×10−6∣∣∣∣∣=4π∈0(10×10−2)23×10−6N/C along OB ∴E=E1+E2 =2×[(9×109)×(10×10−2)23×10−6] [As E1 = E2, the value is multiplied with 2] =5.4×106 N/ C along OB Therefore, the electric field at mid-point O is 5.4×106 N/ C along OB. (b) A test charge of amount q=1.5×10−9C is placed at mid-point O. q=1.5×10−9C Force experienced by the test charge =F ∴F=qE =1.5×10−9×5.4×106 =8.1×10−3N The force is along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A. Hence, the force experienced by the test charge is 8.1×10−3Nalong OA.
